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2 years ago

The expression below is a factorization of what trimonial (5x+6)(3x-2)

Mathematics

2 answers:

9966 [12]2 years ago

5 0

Answer:

15x² + 8x - 12

Step-by-step explanation:

(5x + 6)(3x - 2)

Each term in the second factor is multiplied by each term in the first factor, that is

5x(3x - 2) + 6(3x - 2) ← distribute both parenthesis

= 15x² - 10x + 18x - 12 ← collect like terms

= 15x² + 8x - 12

STatiana [176]2 years ago

3 0

Answer:

$15x^2 + 8x -12$

Step-by-step explanation:

(5x+6)(3x-2)

"foil" it out

F: 5x * 3x = 15x^2

O: 5x * -2 = -10x

I: 6*3x = 18x

L: 6 * -2 = -12

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Find the integration of (1-cos2x)/(1+cos2x)

slega [8]

Given:

The expression is:

$\dfrac{1-\cos 2x}{1+\cos 2x}$

To find:

The integration of the given expression.

Solution:

We need to find the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ .

Let us consider,

$I=\int \dfrac{1-\cos 2x}{1+\cos 2x}dx$

$I=\int \dfrac{2\sin^2x}{2\cos^2x}dx$ $[\because 1+\cos 2x=2\cos^2x,1-\cos 2x=2\sin^2x]$

$I=\int \dfrac{\sin^2x}{\cos^2x}dx$

$I=\int \tan^2xdx$ $\left[\because \tan \theta =\dfrac{\sin \theta}{\cos \theta}\right]$

It can be written as:

$I=\int (\sec^2x-1)dx$ $[\because 1+\tan^2 \theta =\sec^2 \theta]$

$I=\int \sec^2xdx-\int 1dx$

$I=\tan x-x+C$

Therefore, the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ is $I=\tan x-x+C$ .

8 0

2 years ago

Riley has collected surveys for her final thesis project. One one question, 46.1% of respondents said that conditions were impro

ira [324]

Answer:

The maximum value of the confidence interval for this set of survey results is 51.73%.

Step-by-step explanation:

A confidence interval has two bounds, a lower bound and an upper bound.

These bounds depend on the sample proportion and on the margin of error.

The lower bound is the sample proportion subtracted by the margin of error.

The upper bound is the margin of error added to the sample proportion.

In this question:

Sample proportion: 46.1%

Margin of error: 5.63%.

Maximum value is the upper bound:

46.1+5.63 = 51.73

The maximum value of the confidence interval for this set of survey results is 51.73%.

5 0

3 years ago

Please help! This is the third time I asked

Alexxx [7]

Hi,
To find K you will pick a value from the x and then the corresponding Y value. Then we will plug it into the equation that is given y=Kx

Here:
X=25 Y=160
160=k25
(Divide both sides by 25 to isolate K)
K=6.5

Then plug K value into equation
Y=6.5X

Then if you want to check if that is correct you can pick a different point on the table and plug that into the equation to see if you get the same value.

Check:
X=50

Y=6.5x50
Y=320

5 0

2 years ago

I need help with this !!

zubka84 [21]

Answer:

is there more to the problem

5 0

3 years ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

2 years ago